Question

我正在尝试创建一个程序作为练习，其中时钟每秒钟滴答。我试图找到有关这方面的信息，但它对我学到的东西来说太复杂了，或者说不相关。

我希望它能像这样工作：

Clock starts at time: 12:00:00 AM
Clock has been set to time: 11:59:00 PM
TICK: 11:59:01 PM

这是我到目前为止编写的代码：

public class Clock {

    public static void main(String[] args) {
        SimpleClock clock = new SimpleClock();
        System.out.println("Clock starts at time: " + clock.time());
        clock.set(11, 59, 00, false);
        System.out.println("Clock has been set to time: " + clock.time());
        for (int j = 0; j < 60; j++) {
            for (int i = 0; i < 60; i++) {
                clock.tick();
                System.out.println("TICK: " + clock.time());
            }
        }
        System.out.println("Clock finally reads: " + clock.time());
    }

}

GUI：

public class ClockView extends JFrame {

    /* -----------------Private Member variables --------------------- */
    private static final long serialVersionUID = 1L;
    private static int ROWS_IN_GRID = 2;
    private static int COLS_IN_GRID = 1;
    private static int BUTTON_ROWS = 1;
    private static int BUTTON_COLS = 2;
    private SimpleClock clock;
    private JLabel face;

    /**
     * Constructor. Takes a SimpleClock as an argument and builds a graphical
     * interface using that clock the model. The Tick button increments the
     * clock by 1 second, while the Reset button sets the clock back to midnight
     * (12:00:00AM). *
     * 
     * @param clock
     *            - the clock instance used to store the time for the view
     */
    public ClockView(SimpleClock clock) {
        super("SimpleClock Demo");
        this.clock = clock;
        this.face = new JLabel("<html><span style='font-size:20px'>"
                + this.clock.time() + "</span></html>");

        this.setLayout(new GridLayout(ROWS_IN_GRID, COLS_IN_GRID));
        this.add(this.face);

        JPanel buttonPanel = new JPanel(
                new GridLayout(BUTTON_ROWS, BUTTON_COLS));
        JButton tickButton = new JButton("Tick");
        tickButton.addActionListener(new ActionListener() {

            @Override
            public void actionPerformed(ActionEvent arg0) {
                clock.tick();
                ClockView.this.face
                        .setText("<html><span style='font-size:20px'>"
                                + clock.time() + "</span></html>");

            }

        });
        buttonPanel.add(tickButton);

        JButton resetButton = new JButton("reset");
        resetButton.addActionListener(new ActionListener() {

            @Override
            public void actionPerformed(ActionEvent e) {
                clock.set(12, 0, 0, true);
                ClockView.this.face
                        .setText("<html><span style='font-size:20px'>"
                                + clock.time() + "</span></html>");
            }

        });
        buttonPanel.add(resetButton);

        this.add(buttonPanel);
        this.pack();
        this.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        this.setVisible(true);
    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        ClockView v = new ClockView(new SimpleClock());

    }

}

我退出肯定我的逻辑错误发生在public void tick()类<{1}}

中

基本上，它只是从am切换到pm，因为我目前在while循环结束时切换程序。我知道我必须移动它，但我不知道如何，因为时钟甚至不会在第一时间打勾。

Answer 1

您当前的tick方法包含大量循环，最终会以时间＆gt; = 12小时结束。为什么？因为您将添加一秒，直到循环条件不满足。因此，在hours >= 12之前，还要等到minutes = 59或seconds = 59

你应该只添加一秒，就像评论所说：将时钟提前1秒。。然后进行具体检查。

public void tick(){
    seconds++;
    if(seconds => 60){
         seconds = 0;
         minutes++;
         if(....)
             ...
    }
}

此外，morning将始终在此处切换，但您应该只在小时数达到限制时执行此操作

java-在不使用内置时钟类的情况下创建时钟

1 个答案: